Homework5 - P1241 Rfipreaen’rfilg 1250‘ using...

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Unformatted text preview: P1241 Rfipreaen’rfilg 1250‘) using equations: of tha atraight lina iEgfllEfltS gives. {Zr 21):: UEE KCL To get 0 :59: _‘r 20+“: ‘0 flfiirglfl _1d I 1r'5(f]_ 5 I(r)_7E“m+ a _ aim—"rm * —1{J+ q“ lflflrizfl 0 ram P1242 —2r = e A for I :5 [3 Apply KC L To get —2: :(r)=2—1C[r)=2—e A for£>0 1312—13 209 Va 4009 We‘ll write and solve a node equation. Label + the node voltages as shown. Apply KCL at v“) 2 ml: Hm g; o |2 V node a to get 1[r)—1=3 La v1 —12 201(t)+12 : + :3 1:3 :— 20 100 400 25 ‘30 vi =2.4+4.4ae‘5' V for 3:50 Then 3(3): 1’" =24+44.se'5’ 11131 for we Inn PIS—5 Max. charge on capacitor = Cv =(lflx1tT5) (6) = 60 pt: _6 a: = fl = fl = 6 sec to charge 3' llflxlfl 1 g _ 1 _5 1 _ atored energyr — W — ECL — 2(10x1fl —18D J33] PIS—IS We have 1=l[t]+) vflT} = 3 V 12(3) = grimy: +vr({])= Sjgficfl dr + 3 = 3 [e5'—1)+3 = 3 RV. 0:343 a) 1:(3)=vfl[r)+vr(r)= 53(I)+1=fi(t)=159fl+ 3 3’ =13 315’ v. De 3 <1 W(r}|r=fi.2:: L65 J 1: W r =icl=2 r =i 0.2 3.5”:0910’3 2» :l [) 2 ‘-"[) 2X ( e ] 3 Flu-3:: P144 Replacing series and parallel capacitors by equivalent capacitors. the circuit can be reduced as follows: C l (:7 (fl—AL W) C 3C vtr} Fm C 2 C Then at) = %%v z] =§ i4 eos(3r) = %[—12sm(3a]] = —%afl(3:] v 1114-5 The 16 F capacitor is in series with a parallel combination of 4 F and 12 F capacitors. The capacitance of the equivalent capacitor is 16(4+12] —=3 F 16+(4+12] The 30 F capacitor is in parallel with a short circuit. which is equivalent to a short circuit. After making these simplifications. we have 11 C 12F 8F BHM 10F Ceq Then 10(12+C+a) 3=C — 2:: [7:20]: “1 _10+(12+C+8) PTA—6 l C = =1DF eq 1 1 1 1 : + + 60 15+1C| 30 4D+60 (Checked using LNAP 6526.504} P T . 4— T Fir-st Then I C | “a! 1 2 2 C c —- gt“ E+§+§ | C1 | PIS—6 1.11 general m) = (2x103]:_ (r) + [4x10_3)dif,(r) . t . —3 For O<it<11 #5 £53} = (1] fixiEHi}: 1033‘ :> gifl) = 1x103. Conaequen‘rly x f ‘ v0) = (2x103)[1x103) : + 4x10'5(1x103] = [54106: + 4) v For 14m ‘iirfii 3p": 1'53} = 1 111A 2} “ii 11:} = D . Consequently I v(:} = (2x103}(1x10‘3) + [4x1fl'3)x0 = 2 v Inn—3' 1x10'3 3 — —10 . Cause nenfl ' lxlD‘6_ q 3 1MPfi _ Forfiysfiitfifiys islitj=4><10_3—[ ]I 2} dink):— I . a" When 5gb; <13”: his 3'50?) = —l>-: 10'” and d—iSLE) = U . Conaequently I m) = (2x103)(10-3)=—2 V 1x10‘3 lxlCI—fi ' d When maimigfls is“) = £ )r — 3x1fl'3 2-» {ii—1'56}: 1x103 r m) =(2x103](103I—Sx10—3]+[4xlfl'3)(103)=—12+(2><lflfi)r When Sgra (1 r. t11e11i:(r} = '1] => gilt) = O . C nusequently I vfi‘} = D PIS—T (a) d CI ID <:: I< 2 vU) = Ld—m‘) = 0.1 2 < H 6 3‘ CI 6 t: r “3) :(r) = 50111) (27-5 “(0) = 2fiy-(r) dr Fo1‘0‘it<12.v[r)=fl V30 i(t]=2j;0 dr+fl=0 A Fit-1‘2 ‘i H1 6. vhf) = {1.2 f— 0.4 V so in) = #102144) d-I +0 =(«t12r2 413-5) 2:43.29 —U.8r+{].8 A :15) =U.2(6-’)—U.3[5)+U.3 =32 A. Far 6 *i t. wit} = 0.3 V 30 :(r): 2!;03 dr+3.2=(l.6r—6.4] A PIE—4 Fifi—5 di 1' d -1 _I _, v = L— = [1] EH”; ) = (1—0.9 1- P = vi = [ (14);?" ] (42‘3") = 4r(1—r)e‘2r w W =13 = l[l](4re_’)z =2:% '3’ .T "J 7 4 — .— 1- [ {2: ma 0 - - 22‘ {m #1 1 = L£=lfl and i(r)=- a :> 1:“): dr 2dr —2(r—2) 1<H2 —1 +1) r}: 0 I1] H0 _ 2 r {){Hl PU] _ WHO _ ' 23-2] 1<r<2 D r}: W(r) = W(rfl)+j:np[r)dr EU} = [J for r<U := p0] = CI fm' H0 := W[rfl)=fl U<r<1:W[r) =J12rdr = :3 1<r<2 : WU) = W[1)+J':2(r—2)dr = r3 — 4: + 4 wz: WU) = W(2) = a MD {hr-{1 1<r<l ml PIT—4 L 2 K +L]><L . . . +-L £1 The equivalent inductance 15: +2L= L Lle 3 +1, +L L+2L Then i(r)= 1 [I 4eoa[3r] dc: 3 Xian1(3r)=127’ain(3r) 111.551 Elf—r 21x4 - 8 P115 The 25 H inductor ii in r:.erie5 with a parallel combination of 20 H and 60 H inductors. The inductance of the equivalent inductor is 60x20 60+20 25+ =40H The 30 H inductor is in parallel with a short circuit. which is equivalent to a rLllcart circuit. After making these simplifications. we have fit 20 H 40 H 1 O H B Lek-I Then 18=L =10+ 1 2) i+i+i=l 2) L=EDH Eq 1 1 1 20 L 40 3 —+—+— 20 L 40 (Checked using LNAP 6.525.504} P116 15 10 40 60 L —60: K :30: x —60:6+30:24—120H “4 15+1o 40+60 P11? First Then 2 ' 2 12=Leq= .' ' =—L :5 L235 nLH 2 2 55 Lek-I ' ' (Checked using LNAP 6.526.504} PIT—8 . _ 1 3 —5r _ 1 —5r _ -52‘ (a) Elm—E €49 dr+0—_—m[e —l]—O.l(l—€ )A for :20 (b) When!=0.25. t1{0.2)=0.1(1—e'1)= 63.2 111A and i2[0.2]=0.4[1—e_1)=252.8 111A so the energy stored by the S H inductor is wl = %[S)U.06322 = lISJD ml and the energy stored by the 2 H indietor i5 W2 2 63 .9 ml . () L 8'2 16H C = = . 6” 8+2 1 :‘(r]=— r4e‘5mF1r:t::.5(1—.c3'5‘) A f01'r33'D 1.5 n (d) "When I = 0.2 5. 3' (0.2] = {150—94) = 3 16 111A so the energyr attired by the eqlnvalent inductor is w =%{1.6]D.3162 = 79.9 ml = 1121+w2. ...
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This note was uploaded on 01/24/2010 for the course ECE 2025 taught by Professor Juang during the Spring '08 term at Georgia Tech.

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Homework5 - P1241 Rfipreaen’rfilg 1250‘ using...

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